Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 100;
int l[N], r[N];
int sum[N];
double get(int sy, int nx, int ny, int xx) {
  return xx - (xx - nx) * 1.0 / ny * sy;
}
int main() {
  int sy, a, b;
  scanf("%d %d %d", &sy, &a, &b);
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d %d", &l[i], &r[i]);
  for (int i = 1; i <= n; i++) {
    sum[i] = sum[i - 1] + r[i] - l[i];
  }
  int qs;
  scanf("%d", &qs);
  for (int i = 1; i <= qs; i++) {
    int nx, ny;
    scanf("%d %d", &nx, &ny);
    int ls = 0, rs = n + 1;
    while (rs - ls > 1) {
      int mid = ls + rs >> 1;
      if (get(sy, nx, ny, l[mid]) < a)
        ls = mid;
      else
        rs = mid;
    }
    int L = rs;
    ls = 0, rs = n + 1;
    while (rs - ls > 1) {
      int mid = ls + rs >> 1;
      if (get(sy, nx, ny, r[mid]) > b)
        rs = mid;
      else
        ls = mid;
    }
    int R = ls;
    double ans = 0;
    if (L <= R) {
      ans = (sum[R] - sum[L - 1]) * 1.0 / ny * (ny - sy);
    }
    double al = get(sy, nx, ny, l[L - 1]), ar = get(sy, nx, ny, r[L - 1]);
    if (L > 1 && ar >= a) {
      al = max(al, (double)a);
      ar = min(ar, (double)b);
      if (al <= ar) ans += ar - al;
    }
    if (R < n && L - 1 != R + 1) {
      al = get(sy, nx, ny, l[R + 1]), ar = get(sy, nx, ny, r[R + 1]);
      al = max(al, (double)a);
      ar = min(ar, (double)b);
      if (al < ar) ans += ar - al;
    }
    printf("%.10f\n", ans);
  }
  return 0;
}